Find the longest valid obstacle course at each position

Given an array of integers, find the longest non-decreasing subsequence. This problem is very similar to Longest increasing subsequence. We will solve this with a greedy approach. The key is, the longest course for index i is determined by two factors:

• obstacles[i], which is required.
• the longest course before index i whose last obstacle is shorter than or equal to obstacles[i].

By combining the two terms above, we can determine the longest course for index i.

In short, the longest course ending at index i depends on the courses ending before index i.

class Solution:
    def longestObstacleCourseAtEachPosition(self, obstacles: List[int]) -> List[int]:
        res = []
        courses = []
        for obstacle in obstacles:
            if not courses or obstacle >= courses[-1]:
                courses.append(obstacle)
                res.append(len(courses))
            else:
                index = bisect.bisect_right(courses, obstacle)
                courses[index] = obstacle
                res.append(index + 1)
        return res

Time complexity: O(nlogn)
Space complexity: O(n)